Notice how the integrand and limits of integration are changed. The change is similar for non-rectangular regions, since they do not change the \(z\) coordinates.
Activity63.
The purpose of this activity is to derive the form for a triple integral in cylindrical coordinates. It is achieved by applying the idea from double to triple in Cartesian to the double integral in cylidrical.
For double integrals in cylindrical coordinates the integral (over a circular region) looks like
\begin{equation*}
\int_a^b \int_c^d z(r,\theta) r \; dr \: d\theta.
\end{equation*}
(a)
Write the general form of the triple integral for the volume given by the double integral above.
Because this is single simple, surface we can integrate in any order. We choose here to integrate in the order \(r, \theta, z\text{.}\) Also, because it is a single surface we start at \(r=0\) and integrate to the surface.
\begin{equation*}
\int_0^{(3-z)(1+\sin\theta)} r \; dr
\end{equation*}
There are no restrictions on \(\theta\) nor interaction with \(z\text{.}\) This means we integrate
\begin{equation*}
\int_0^{2\pi} \int_0^{(3-z)(1+\sin\theta)} r \; dr \; d\theta
\end{equation*}
Vertically the bottom is the xy-plane and the top is the highest value of the surface. Not at \(z=3\)\(r=0\text{,}\) that is, the surface pinches off. The equation produces a surface above \(z=3\text{,}\) but that is not between the surface and the xy-plane. Thus the final integral is
\begin{equation*}
\int_0^3 \int_0^{2\pi} \int_0^{(3-z)(1+\sin\theta)} r \; dr \; d\theta \; dz
\end{equation*}
ExercisesExercises
1.
Find the volume enclosed between \(z=1-r\) and the \(xy\)-plane using a triple integral.
2.
Find the volume enclosed between \(z=1+\cos\theta\) and the \(xy\)-plane using a triple integral.
3.
Find the volume enclosed between \(z=1-r\) and the \(xy\)-plane over the region \(r=\cos\theta\) using a triple integral.
